Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate discrete distributions ▷ Binomial distribution ▷ Conditional binomial

Theorem: Let $X$ and $Y$ be two random variables where $Y$ is binomially distributed conditional on $X$

$\label{eq:Y-X-dist} Y \vert X \sim \mathrm{Bin}(X, q)$

and $X$ also follows a binomial distribution, but with different success frequency:

$\label{eq:X-dist} X \sim \mathrm{Bin}(n, p) \; .$

Then, the maginal distribution of $Y$ unconditional on $X$ is again a binomial distribution:

$\label{eq:Y-dist} Y \sim \mathrm{Bin}(n, pq) \; .$

Proof: We are interested in the probability that $Y$ equals a number $m$. According to the law of marginal probability or the law of total probability, this probability can be expressed as:

$\label{eq:Y-dist-s1} \mathrm{Pr}(Y = m) = \sum_{k=0}^{\infty} \mathrm{Pr}(Y = m \vert X = k) \cdot \mathrm{Pr}(X = k) \; .$

Since, by definitions \eqref{eq:X-dist} and \eqref{eq:Y-X-dist}, $\mathrm{Pr}(X = k) = 0$ when $k > n$ and $\mathrm{Pr}(Y = m \vert X = k) = 0$ when $k < m$, we have:

$\label{eq:Y-dist-s2} \mathrm{Pr}(Y = m) = \sum_{k=m}^{n} \mathrm{Pr}(Y = m \vert X = k) \cdot \mathrm{Pr}(X = k) \; .$

Now we can take the probability mass function of the binomial distribution and plug it in for the terms in the sum of \eqref{eq:Y-dist-s2} to get:

$\label{eq:Y-dist-s3} \mathrm{Pr}(Y = m) = \sum_{k=m}^{n} {k \choose m} \, q^m \, (1-q)^{k-m} \cdot {n \choose k} \, p^k \, (1-p)^{n-k} \; .$

Applying the binomial coefficient identity ${n \choose k} {k \choose m} = {n \choose m} {n-m \choose k-m}$ and rearranging the terms, we have:

$\label{eq:Y-dist-s4} \mathrm{Pr}(Y = m) = \sum_{k=m}^{n} {n \choose m} \, {n-m \choose k-m} \, p^k \, q^m \, (1-p)^{n-k} \, (1-q)^{k-m} \; .$

Now we partition $p^k = p^m \cdot p^{k-m}$ and pull all terms dependent on $k$ out of the sum:

$\label{eq:Y-dist-s5} \begin{split} \mathrm{Pr}(Y = m) &= {n \choose m} \, p^m \, q^m \sum_{k=m}^{n} {n-m \choose k-m} \, p^{k-m} \, (1-p)^{n-k} \, (1-q)^{k-m} \\ &= {n \choose m} \, (p q)^m \sum_{k=m}^{n} {n-m \choose k-m} \, \left( p (1-q) \right)^{k-m} \, (1-p)^{n-k} \; . \end{split}$

Then we subsititute $i = k - m$, such that $k = i + m$:

$\label{eq:Y-dist-s6} \mathrm{Pr}(Y = m) = {n \choose m} \, (p q)^m \sum_{i=0}^{n-m} {n-m \choose i} \, \left( p - pq \right)^{i} \, (1-p)^{n-m-i} \; .$

According to the binomial theorem

$\label{eq:bin-th} (x+y)^n = \sum_{k=0}^{n} {n \choose k} \, x^{n-k} \, y^k \; ,$

the sum in equation \eqref{eq:Y-dist-s6} is equal to

$\label{eq:Y-dist-sum} \sum_{i=0}^{n-m} {n-m \choose i} \, \left( p - pq \right)^{i} \, (1-p)^{n-m-i} = \left( (p-pq)+(1-p) \right)^{n-m} \; .$

Thus, \eqref{eq:Y-dist-s6} develops into

$\label{eq:Y-dist-s7} \begin{split} \mathrm{Pr}(Y = m) &= {n \choose m} \, (p q)^m (p - pq + 1 - p)^{n-m} \\ &= {n \choose m} \, (p q)^m (1 - pq)^{n-m} \end{split}$

which is the probability mass function of the binomial distribution with parameters $n$ and $pq$, such that

$\label{eq:Y-dist-qed} Y \sim \mathrm{Bin}(n, pq) \; .$
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Metadata: ID: P358 | shortcut: bin-margcond | author: JoramSoch | date: 2022-10-07, 21:03.